A refinement of the Gauss-Lucas theorem (after W. P.
Thurston)
Arnaud Chéritat, Yan Gao, Yafei Ou, Lei Tan
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Arnaud Chéritat, Yan Gao, Yafei Ou, Lei Tan. A refinement of the Gauss-Lucas theorem (after
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A refinement of the Gauss-Lucas theorem (after W. P. Thurston)
Un raffinement du théorème de Gauss-Lucas (d’après W. P. Thurston)
Arnaud Chéritat a , Yan Gao b , Yafei Ou c , Lei Tan d
a Insititut
de Mathématiques de Bordeaux, Université de Bordeaux, 351,
cours de la Libération - F 33 405 TALENCE, France,
b Mathematical department of SiChuan University, 610065, ChengDu, China
c SJTU-ParisTech Elite Institute of Technology, 200240, ShangHai, China
d Université d’Angers Faculté des sciences LAREMA, 49045, Angers, France
Received *****; accepted after revision +++++
Presented by
Abstract
We present a refinement of the Gauss-Lucas theorem, following an idea of W. P. Thurston.
Résumé
Nous présentons un raffinement du Théorème de Gauss-Lucas, d’après une idée de W. P. Thurston.
1. Introduction
This article, although self contained, can be considered as a completion of the article [1], where the
statements and ideas of proofs have been exposed.
Let f be a polynomial with complex coefficients of degree m ≥ 1. By the fundamental theorem of
algebra, f has exactly m complex roots counting multiplicities. The derivative f 0 is a polynomial of
degree m − 1 and its roots of are called the critical points of f .
The following theorem by Gauss and rediscovered by Lucas describes a beautiful relationship between
the roots of a polynomial and the roots of its derivative.
Email addresses: arnaud.cheritat@math.univ-toulouse.fr (Arnaud Chéritat), gyan@scu.edu.cn (Yan Gao),
oyafei@sjtu.edu.cn (Yafei Ou), tanlei@math.univ-angers.fr (Lei Tan).
Preprint submitted to the Académie des sciences
April 23, 2015
Theorem 1.1 (Classical version of the Gauss-Lucas theorem) If f is a polynomial of degree at
least two, then the convex hull of the roots of f contains the roots of f 0 .
The proof of this theorem is fairly elementary and can be found in many text books. W. Thurston gave
a purely geometric proof of the theorem eliminating all the calculations except the fact that a polynomial
is a product of linear factors. We present his proof in Section 2. He also showed another version of this
theorem which goes as follows:
Proposition 1.1 (A surjective version of the Gauss-Lucas theorem) Let f be any polynomial of
degree at least two. Denote by C the convex hull of the critical points of f . Then f : E → C is surjective
for any closed half-plane E intersecting C.
This is an equivalent statement of the Gauss-Lucas theorem. In fact, if the convex hull of the roots of
f does not contain C, there would exist a closed half-plane E intersecting C but avoiding all roots of f .
Hence f |E would be non surjective and hence Proposition 1.1 implies Theorem 1.1. Conversely, one sees
that Theorem 1.1 implies Proposition 1.1 by considering the convex hull of roots of f − w for all complex
constants w.
By this proposition, if we want to cover the whole plane with a half-plane containing the least possible
points, we should take the one just touching C. To achieve this covering more efficiently, one may look
for a connected region D such that f is surjective on D and injective on the interior of D. Thurston, in
private communications, indicated to us the following result, which is more powerful than the classical
Gauss-Lucas theorem:
Theorem 1.2 (The Gauss-Lucas-Thurston theorem) Let f be any polynomial of degree at least
two, and denote by C the convex hull of the critical points of f . Let L be a straight line intersecting C
and which bounds an open half plane H disjoint from C. If c ∈ L is a critical point of f , then there exists
a set of geodesics for the metric |f 0 (z)| · |dz| starting at c and with directions in H that forms an open
subset U of H for which f (U ) = C. Furthermore, f maps U bijectively onto C minus a radial line from
f (c) to ∞ (see Figure 1).
H
U
L
f
bb
c1
b
f (c1 )
c2
b
C
b
c4
b
c3
Figure 1. Bijective region of f
We had never seen such statement elsewhere. We have completed a proof of it following the guide lines
of Thurston. The proof will be presented in Section 3.
2
2. A geometric proof of Gauss-Lucas theorem, due to Thurston
Proof. Let a ∈ C. Consider the distance function z 7→ |z − a|. The level curves of this function are
concentric circles with center a, so the gradient direction of the function at any point z 6= a is the radial
direction from a to z (see Figure 2). Let L be the line through z and perpendicular to the segment [a, z].
L3
β
b
z
gradient direction at z of |z − a|
b
z
α
H
b
L2
a4
bb
a1
Z
bb
b
a
a3
L4
L
b
a2
Figure 2. Gradient direction of z 7→ |z − a|
L1
Figure 3. The intersection of all Hj
Denote by H the open half-plane determined by L disjoint from a. Then the function z 7→ |z − a| has a
strictly positive directional derivation at z in any direction pointing into H.
Let f be a polynomial of degree at least two. Denote by Z the convex hull of the roots of f and let z
be a point outside Z. For each root aj of f , we can draw a line Lj through z and perpendicular to the
segment [z, aj ]. Denote by Hj the open half-planes determined by Lj disjoint from aj (see Figure 3). By
the argument above, each distance function z 7→ |z − aj | has a strictly positive directional derivation at
z in any direction pointing into Hj .
The fact that z 6∈ Z is equivalent to say that the point z admits a cone of angle α < π containing all
roots of f . Then the intersection of all Hj is a cone based at z with angle β = π − α > 0 (the shadow
region in Figure 3). Therefore, there exists a direction in which the directional derivative of z 7→ |z − aj |
is strictly positive for each root aj of f . We see that, in this direction, the directional derivative of the
function z 7→ |f (z)| is still strictly positive, hence non-zero. Then we conclude that f 0 (z) 6= 0. Otherwise,
if f 0 (z) = 0, the directional derivative of f in each direction at z is 0, so is the one of |f |, which is a
contradiction to the argument above.
3. The proof of Theorem 1.2
Proof. We firstly treat the case of simple critical points, i.e., f 00 (c) 6= 0. In this case, we will show that
the region U in Theorem 1.2 consists of all geodesics starting at c and with directions in H.
We may assume that c = 0, f (0) = f 0 (0) = 0, H = H+ , the upper half-plane, and f 00 (0) = 1. Note that
the conformal metric |f 0 (z)||dz|, which is a Riemannian metric with the singularities at critical points of
f , is obtained by pulling-back the Euclidean metric by f , so the geodesics in this metric are pullback
by f of straight lines. Then we just need to prove that there exist two branches of f −1 (R+ ) emanating
from 0, belonging to H+ := H+ ∪ R, and such that the open region U ⊂ H+ bounded by these two curves
3
contains no curves of f −1 (R+ ). If so, the map f : U → C \ R+ is a covering, and hence must be bijective
by the Riemann-Hurwitz formula.
First, we deal with the case that at least one critical point is in the open lower half plane H− . As 0
is a simple critical point, then there are two branches R1 , R2 in f −1 (R+ ) emanating from 0, and each
component of C \ (R1 ∪ R2 ) has the angle π/2 at 0. We want to show that R1 ∪ R2 ⊂ H+ ∪ {0}.
For this we study the behavior of the vector field 1/f 0 (z) on L = R near 0, because the two curves
∂
.
R1 , R2 are trajectories of the vector field f 01(z) ∂z
Q
Q
z
0
We have f (z) = A · j (cj − z) with A = cj and =(cj ) ≤ 0 for any non-zero critical point cj of f .
Fix x ∈ R close to 0. Write
cj − x = rj (x)eiθj (x) with rj (x) > 0 and 0 ≤ θj (x) < 2π. In particular
Q
iθj (0)
cj = rj (0)e
. So A = rj (0)eiθj (0) . Then, for x 6= 0, there is some r(x) > 0 so that

−1
P
1
A Y
 = r(x) ei (θj (0)−θj (x)) .
=
(c
−
x)
j
f 0 (x)
x
x
j
Note that for x close to 0, the quantity θj (0) − θj (x) is either 0 (cj ∈ R) or close to 0 and has the same
sign as x (cj ∈ H− ) (see Figure 4). Since at least one critical point is in H− , then = f 01(x) > 0 for any x 6= 0
U
R2
R1
γ2
γ1
U
x′ 0 x
b
b
b
θ1 (x)
θ1 (x′ )
θ1 (0)
b
c3
c1
b
b
b(w) = −2
b
v=
b
5
24
0
b
w=
b
4
3
a(v) = 1
b
c2
Figure 4. The same sign of θ1 (0) − θ1 (x) and x
Figure 5. Bijective region of f =
1 2
z
2
− 61 z 3 − 18 z 4
close to 0. It follows that for z close to 0, the vector 1/f 0 (z) points into H+ . Since the two curves R1 , R2
∂
, according to Gronwall’s comparison lemma, we conclude that
are trajectories of the vector field f 01(z) ∂z
they belong locally to H+ .
We now show that the two trajectories stay entirely in H+ . Let a ∈ H+ be a point of the two trajectories.
Let γ(t) be either R1 or R2 so that f (γ(t)) = t · f (a), t ∈ R+ . Then γ 0 (t) · f 0 (γ(t)) = f (a) and γ 00 (t) ·
f 0 (γ(t)) + γ 0 (t)2 f 00 (γ(t)) ≡ 0. It follows that
X
d
1
γ 00 (t)
f 00 (γ(t))
1
1
(= 0 ) = =(− 0 2 ) = = 0
==
+
=
<0
dt γ (t)
γ (t)
f (γ(t))
γ(t)
γ(t) − cj
j
This means that = γ 01(t) is decreasing and therefore =γ 0 (t) is increasing. Write γ(t) = x(t) + iy(t). We
+
know that y 0 (t) > 0 for small t and y 0 (t) is increasing. So y 0 (t) > 0 for all t ∈ R+
∗ := R \ {0}. It follows
+
that y(t) is also increasing. But y(t) > 0 for t small. So y(t) > 0 for all t ∈ R∗ .
It remains to show that the open region U ⊂ H+ bounded by R1 , R2 contains no branches in f −1 (R+ ).
As no critical points of f are in H+ , the region U \ f −1 (R+ ) is connected and simply-connected, and the
4
map f : U \ f −1 (R+ ) → C \ R+ is bijective by the Riemann-Hurwitz formula. Assume that a branch
R of f −1 (R+ ) belongs to U . Since f (R) = f (R1 ) = R+ , there exists a point in a neighborhood of R+
having two preimages in U \ f −1 (R+ ), which belong to neighborhoods of R and R1 respectively. It is a
contradiction. So we obtain that U ∩ f −1 (R+ ) = ∅.
Now we assume that all critical points of f are in R. Then f is a polynomial with real coefficients.
For each t ∈ R+ close to 0, there is a unique positive (resp. negative) number a(t) (resp. b(t)) such that
f (a(t)) = t (resp. f (b(t)) = t). The function a(t) (resp. b(t)) is increasing (resp. decreasing) in the interval
(0, v) (resp. (0, w)), where a(v) (resp. b(w)) is the minimal positive (resp. maximal negative) critical point
of f . Suppose the multiplicity as a critical point of a(v) (resp. b(w)) is k ≥ 1, then there is a curve γ1 in
f −1 [v, +∞) (resp. γ2 in f −1 [w, +∞)) such that the angle between γ1 and [0, a(v)] at a(v) (resp. γ2 and
π
and γ1 (resp. γ2 ) belongs locally to H+ . With the same argument as in the first
[b(w), 0] at b(w)) is k+1
case, we get that γ1 and γ2 stay entirely in H+ (see Figure 5).
Set R1 := [0, a(v)] ∪ γ1 and R2 := [b(w), 0] ∪ γ2 . Under our construction, the open region U ⊂ H+
bounded by R1 and R2 is not disconnected by f −1 (R+ ). So we have U ∩ f −1 (R+) = ∅ by the same reason
as the first case. Then the proof to the case of simple critical points is completed.
For the case of multiple critical points, if c ∈ L is a critical point of f with multiplicity k ≥ 2, we
2π
at c forms an region U
conclude that a set of geodesics from c with directions in a cone in H of angle k+1
with the properties in Theorem 1.2. The proof is very similar as the case of simple critical points, where
the local part is somehow simpler and the global part is completely the same.
Remark 1 In fact, it is not difficult to find the bijective regions, and we have a lot of choices for the image
of their interior by f . The interest of this theorem is that one of the bijective regions is in the half-plane
H supported by C.
References
[1] A. Chéritat A, L. Tan, Si nous faisons danser les racines? Un hommage à Bill Thurston, article de vulgarisation, Images
des mathématiques CNRS(2012).
5